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Chapter 3: Problem 6

Find the first and second derivatives. $$y=\frac{x^{3}}{3}+\frac{x^{2}}{2}+\frac{x}{4}$$

Short Answer

Expert verified
First derivative: \( x^2 + x + \frac{1}{4} \); Second derivative: \( 2x + 1 \).

Step by step solution

01

Recall the Power Rule for Differentiation

The power rule states that if you have a term of the form \( ax^n \), its derivative is \( a \cdot n \cdot x^{n-1} \). This rule will be used to differentiate each term of the function \( y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4} \).
02

Differentiate each term to find the first derivative

Apply the power rule to each term:- The derivative of \( \frac{x^3}{3} \) is \( 3 \cdot \frac{1}{3} \cdot x^{3-1} = x^2 \).- The derivative of \( \frac{x^2}{2} \) is \( 2 \cdot \frac{1}{2} \cdot x^{2-1} = x \).- The derivative of \( \frac{x}{4} \) is \( \frac{1}{4} \cdot 1 \cdot x^{1-1} = \frac{1}{4} \).Therefore, the first derivative \( y' \) is \( y' = x^2 + x + \frac{1}{4} \).
03

Differentiate the first derivative to find the second derivative

Now differentiate \( y' = x^2 + x + \frac{1}{4} \) similarly to find \( y'' \):- The derivative of \( x^2 \) is \( 2 \cdot x^{2-1} = 2x \).- The derivative of \( x \) is \( 1 \cdot x^{1-1} = 1 \).- The derivative of the constant \( \frac{1}{4} \) is 0.Thus, the second derivative \( y'' \) is \( y'' = 2x + 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The Power Rule is a fundamental concept in differentiation, used to find the derivative of a term that is in the form of \( ax^n \). It is incredibly useful for quickly and easily determining how a function changes. To apply the power rule, simply follow these steps:
  • Multiply the coefficient \( a \) by the exponent \( n \).
  • Reduce the exponent by one to find the new power of \( x \).
  • Combine these pieces to form the derivative.
For example, to differentiate \( \frac{x^3}{3} \), use the power rule by doing the following: 1. Multiply the 3 (power) by \( \frac{1}{3} \) (coefficient). 2. Lower the power from 3 to 2.So, the derivative becomes \( x^2 \). The power rule is straightforward, but it's key to mastering differentiation, making it a jewel for tackling polynomial functions.
First Derivative
The first derivative of a function provides valuable information about the rate of change of the function. It is also fundamental in understanding the slope of the tangent line at any point on the curve.
To find the first derivative of the function \( y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4} \), apply the power rule to each term:
  • \( \frac{x^3}{3} \) becomes \( x^2 \)
  • \( \frac{x^2}{2} \) becomes \( x \)
  • \( \frac{x}{4} \) becomes \( \frac{1}{4} \)
Thus, the first derivative is \( y' = x^2 + x + \frac{1}{4} \).
This first derivative tells us how the values of \( y \) change based on changes in \( x \). Think of it as a tool to measure how steep or flat the function curve is.
Second Derivative
The second derivative of a function provides deeper insight into the nature of the function's graph. It reveals the rate at which the first derivative itself is changing.
To find the second derivative from the first derivative \( y' = x^2 + x + \frac{1}{4} \), apply the power rule again to each term:
  • The derivative of \( x^2 \) becomes \( 2x \)
  • The derivative of \( x \) becomes 1
  • The derivative of the constant \( \frac{1}{4} \) is 0
Thus, the second derivative is \( y'' = 2x + 1 \).
The second derivative can tell us about concavity and inflection points of the graph. If the second derivative is positive, the graph is concave up (like a smile), and if negative, it's concave down (like a frown). Knowing how the graph curves enables better understanding of the behavior of the function in question.

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Most popular questions from this chapter

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$

a. Find the tangent to the curve \(y=2 \tan (\pi x / 4)\) at \(x=1\) b. What is the smallest value the slope of the curve can ever have on the interval \(-2

Using the Chain Rule, show that the Power Rule \((d / d x) x^{n}=n x^{n-1}\) holds for the functions \(x^{n}\). $$x^{1 / 4}=\sqrt{\sqrt{x}}$$

Find \(y^{\prime \prime}\) $$y=9 \tan \left(\frac{x}{3}\right)$$

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